挑战程序竞赛系列（17）：3.1最大化平均值

详细代码可以fork下Github上leetcode项目，不定期更新。

练习题如下：

• POJ 2976: Dropping tests
• POJ 3111: K Best

POJ 2976: Dropping tests

01分数规划，具体可以参考博文http://www.cnblogs.com/perseawe/archive/2012/05/03/01fsgh.html，写的非常详细且通俗易懂。

我的认识：中学里学过二分法求函数零点，其实该代码模拟的就是这种求解过程，所以只要写出目标函数，零点不难求。

接着就得到了求和式：

∑i=1n(ai−l⋅bi)≥0

要删除k个物品，当然选择贡献最小的，所以按照ai−l⋅bi排序，选择n-k个物品，排除其余贡献较小的k个物品。

代码如下：

static class Pair implements Comparable<Pair>{        int a;        int b;        @Override        public int compareTo(Pair o) {            double ocmp = o.a - L * o.b;            double tcmp = this.a - L * this.b;            return ocmp == tcmp ? 0 : (ocmp > tcmp) ? 1 : -1;        }    }    static Pair[] p;    static double L;    static int k;    static int n;    public static void main(String[] args) throws IOException {        Scanner in = new Scanner(System.in);        while (true){            n = in.nextInt();            k = in.nextInt();            if (n == 0 && k == 0) break;            p = new Pair[n];            for (int i = 0; i < n; ++i){                p[i] = new Pair();                p[i].a = in.nextInt();            }            for (int i = 0; i < n; ++i){                p[i].b = in.nextInt();            }            double lf = 0;            double rt = 1;            while (Math.abs(rt -lf) > 1e-4){                double mid = (lf + rt) / 2;                if (!valid(mid)) rt = mid;                else lf = mid;            }            System.out.printf("%.0f\n",lf * 100);        }    }    public static boolean valid(double mid){        L = mid;        Arrays.sort(p);        double sum = 0;        for (int i = 0; i < n - k; ++i){            sum += (p[i].a - L * p[i].b);        }        return sum > 0;    }

POJ 3111: K Best

和上一题一个思路，但此题精度要求要高很多，按照上一题的while循环，WA了，测了下，大概循环14次左右，而此题循环50次，可以达到0.000000001的精度。注意记录下下标。

代码如下：

static class Pair implements Comparable<Pair>{        int a;        int b;        int id;        @Override        public int compareTo(Pair o) {            double that = o.a - L * o.b;            double thiz = this.a - L * this.b;            return that == thiz ? 0 : (that > thiz) ? 1 : -1;        }    }    static double L;    static Pair[] p;    static int N;    static int K;    public static void main(String[] args) throws IOException {        Scanner in = new Scanner(System.in);        N = in.nextInt();        K = in.nextInt();        p = new Pair[N];        ans = new int[K];        for (int i = 0; i < N; ++i) p[i] = new Pair();        double max = 0;        for (int i = 0; i < N; ++i){            p[i].a = in.nextInt();            p[i].b = in.nextInt();            p[i].id = i + 1;            max = Math.max(max, p[i].a / (p[i].b * 1.0 ));        }        double lf = 0.0;        double rt = max;        for (int i = 0; i < 50; ++i){            double mid = (lf + rt) / 2;            if (!valid(mid)) rt = mid;            else lf = mid;        }        for (int i = 0; i < K; ++i) ans[i] = p[i].id;        StringBuilder sb = new StringBuilder();        Arrays.sort(ans);        for (int i : ans){            sb.append(i + " ");        }        System.out.println(sb.toString().trim());    }    static int[] ans;    public static boolean valid(double mid){        L = mid;        Arrays.sort(p);        double sum = 0.0;        for (int i = 0; i < K; ++i){            sum += (p[i].a - L * p[i].b);        }        return sum > 0;    }